# Difference between revisions of "1987 IMO Problems/Problem 5"

## Problem

Let $n$ be an integer greater than or equal to 3. Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.

## Solution

Consider the set of points $S = \{ (x,x^2) \mid 1 \le x \le n , x \in \mathbb{N} \}$ in the $xy$-plane.

The distance between any two distinct points $(x_1,x^2_1)$ and $(x_2,x^2_2)$ in $S$ (with $x_1 \neq x_2$) is:

$d = \sqrt{(x_1-x_2)^2+\left(x^2_1-x^2_2\right)^2} =$ $\sqrt{(x_1-x_2)^2+(x_1-x_2)^2(x_1+x_2)^2} = |x_1-x_2|\sqrt{1+(x_1+x_2)^2}$.

Since $1+(x_1+x_2)^2$ is an integer and not a perfect square, $\sqrt{1+(x_1+x_2)^2}$ is irrational. Since $|x_1-x_2|$ is a non-zero integer, $d$ is irrational as desired.

All the points in the set lie on the parabola $y = x^2$. Thus, it is impossible of any set of three points to be collinear, since no line can intersect a parabola at more than two points. Therefore, any triangle with all vertices in $S$ must be non-degenerate as desired.

Since all the points are lattice points, by Pick's Theorem, the area of any triangle with all vertices in $S$ must be in the form $A = I + \dfrac{B}{2} - 1$ where $I$ and $B$ are integers. Thus, the area of the triangle must be rational as desired.

This completes the proof.

 1987 IMO (Problems) • Resources Preceded byProblem 4 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 6 All IMO Problems and Solutions